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Taylor Series and Error Bounds · Tutorial 894 of 999

Approximating Function Values with Taylor Series

Choose an effective Taylor expansion, compute its value at the requested input, and use a rigorous error bound to decide how many decimal places you can report.

Advanced 9 min read

What You'll Learn

  • Select a convenient center or substitution for a specified function value
  • Evaluate Taylor polynomials while keeping intermediate values visible
  • Apply Lagrange and alternating-series remainder bounds to numerical approximations
  • Use a logarithmic power series and a geometric tail estimate
  • Check whether an error interval supports the decimal places you report
  • Recognize common mistakes in selecting terms and interpreting error bounds

From a Series to a Numerical Value

A Taylor series describes a function using an infinite sum, but an approximation uses only finitely many terms. To approximate a requested function value well, you need to make two decisions: which expansion is convenient at that input, and how many terms give an error small enough for the precision you want to report.

The approximation and its error bound play different roles. The polynomial gives a numerical estimate. The bound tells you how far the true value can be from that estimate. A decimal answer is justified only when the bound supports its reported precision.

Definition: If \(P_n(x)\) is a Taylor polynomial used to approximate \(f(x)\), its remainder is \(R_n(x)=f(x)-P_n(x)\). An error bound is a number \(B\) for which \(|R_n(x)|\le B\). Thus \(P_n(x)-B\le f(x)\le P_n(x)+B\).

Choosing a Useful Expansion

For exponential and trigonometric values near zero, Maclaurin series are often convenient. The first terms are

$$ e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!},\qquad \sin x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!},\qquad \cos x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}. $$

For a logarithm, a substitution can make a series converge more quickly than using an expansion whose input is farther from its center. For \(x>0\), set \(z=(x-1)/(x+1)\). When \(|z|<1\), the logarithmic series gives

$$ \ln x=2\left(z+\frac{z^3}{3}+\frac{z^5}{5}+\frac{z^7}{7}+\cdots\right). $$

For example, when \(x=1.2\), this substitution gives \(z=1/11\), a small input. The powers of \(1/11\) decrease quickly, so just a few terms can give a precise value.

Before calculating, identify the input and the expansion that matches it. Then decide what error bound applies. Taylor’s theorem gives the Lagrange bound \( |R_n(x)|\le M|x-a|^{n+1}/(n+1)!\), provided the \((n+1)\)st derivative exists between the center \(a\) and \(x\) and has absolute value at most \(M\) there. For an alternating series with decreasing term magnitudes tending to zero, the error after a partial sum is at most the magnitude of the first omitted term.

Worked Example: Approximating an Exponential Value

Worked Example: Approximating \(e^{0.2}\)

Use the degree-six Maclaurin polynomial and a Lagrange bound. The polynomial is

$$ P_6(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}. $$

Substitute \(x=0.2\), showing each term:

$$ \begin{aligned} P_6(0.2) &=1+0.2+\frac{0.04}{2}+\frac{0.008}{6} +\frac{0.0016}{24}+\frac{0.00032}{120}+\frac{0.000064}{720}\\ &=1+0.2+0.02+0.0013333333+0.0000666667\\ &\quad+0.0000026667+0.0000000889\\ &\approx 1.2214027556. \end{aligned} $$

The seventh derivative of \(e^x\) is \(e^x\). On \(0\le x\le0.2\), it is less than \(3\), so the Lagrange bound is

$$ |R_6(0.2)|\le \frac{3(0.2)^7}{7!} =\frac{3(0.0000128)}{5040} <0.00000000762. $$

Consequently, \(e^{0.2}\) lies between \(1.2214027479\) and \(1.2214027633\). Both endpoints round to \(1.221403\) to six decimal places, so the justified result is \(e^{0.2}\approx1.221403\) to six decimal places. The check is that this reported value differs from the polynomial estimate by about \(0.0000002444\), which is within the permitted error once rounding is included.

Worked Example: Approximating a Sine Value

Worked Example: Approximating \(\sin(0.4)\)

Use the degree-seven Maclaurin polynomial:

$$ P_7(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}. $$

Evaluate every term at \(0.4\):

$$ \begin{aligned} P_7(0.4) &=0.4-\frac{0.064}{6}+\frac{0.01024}{120}-\frac{0.0016384}{5040}\\ &=0.4-0.0106666667+0.0000853333-0.0000003251\\ &\approx0.3894183416. \end{aligned} $$

The sine terms alternate. Their magnitudes decrease here: the ratio of consecutive nonzero magnitudes is \(0.4^2/((2k+2)(2k+3))<1\). Therefore the alternating-series error is no greater than the first omitted term:

$$ |R_7(0.4)|\le\frac{0.4^9}{9!} =\frac{0.000262144}{362880} <0.000000000723. $$

Thus the value lies between \(0.3894183408\) and \(0.3894183424\). Every number in this interval rounds to \(0.38941834\) to eight decimal places. Hence \(\sin(0.4)\approx0.38941834\) to eight decimal places. Substitution into the polynomial gives the stated estimate, and the interval check confirms that the omitted terms cannot change those rounded digits.

Worked Example: Approximating a Logarithm

Worked Example: Approximating \(\ln(1.2)\)

For \(x=1.2\), the substitution gives \(z=(1.2-1)/(1.2+1)=0.2/2.2=1/11\). Keep the first four terms, through \(z^7/7\):

$$ \begin{aligned} S &=2\left(\frac1{11}+\frac1{3\cdot11^3} +\frac1{5\cdot11^5}+\frac1{7\cdot11^7}\right)\\ &=2\left(\frac1{11}+\frac1{3993} +\frac1{805255}+\frac1{136410197}\right)\\ &\approx 0.18232155670. \end{aligned} $$

All omitted terms are positive. For the tail, the denominators \(9,11,13,\ldots\) are at least \(9\). Bounding each one by \(9\) and summing the resulting geometric series gives

$$ \begin{aligned} 0<\ln(1.2)-S &\le \frac{2}{9}\left(z^9+z^{11}+z^{13}+\cdots\right)\\ &=\frac{2z^9}{9(1-z^2)} =\frac{2(1/11)^9}{9(1-1/121)} <0.000000000096. \end{aligned} $$

The logarithm therefore lies in an interval from \(S\) to \(S+0.000000000096\). Its endpoints round to \(0.182322\) to six decimal places, so \(\ln(1.2)\approx0.182322\) to six decimal places. The check is that the positive tail is smaller than the displayed error bound; rounding the series sum is not being mistaken for an exact evaluation.

Why the Error Bounds Work

The Lagrange bound follows directly from Taylor’s theorem: the remainder has the form \(f^{(n+1)}(c)(x-a)^{n+1}/(n+1)!\) for some \(c\) between \(a\) and \(x\). Taking absolute values and using \(|f^{(n+1)}(c)|\le M\) gives the stated bound. The important practical step is to find a valid \(M\) on the whole interval, not just evaluate the derivative at the input.

The logarithmic tail estimate uses a different idea. Once \(0<z<1\), the omitted powers are \(z^9,z^{11},z^{13},\ldots\). Replacing each denominator by the smaller denominator \(9\) makes the estimate larger, and the remaining powers sum to \(z^9/(1-z^2)\). This produces a rigorous upper bound without needing to calculate infinitely many terms.

Both arguments turn a finite calculation into a statement about the true function value. A polynomial by itself gives an estimate; the remainder argument explains what conclusions that estimate supports.

Reporting Precision and Common Pitfalls

To justify rounding to \(d\) decimal places, it is not enough to say the error is small. Use the error bound to form an interval containing the true value, then check that the entire interval rounds to the same \(d\)-place number. The examples above use this test explicitly.

Common mistake: Claiming that an approximation is correct to six decimal places merely because its error is less than \(0.000001\). That bound can still cross a rounding threshold. For example, a value near \(2.345678\) with error at most \(0.0000008\) could lie on either side of the six-place rounding boundary \(2.3456785\). Form the interval and check its endpoints before reporting the digits.

Other errors to avoid include using the first omitted term as a bound when the series is not alternating with decreasing term magnitudes, and applying a Lagrange bound without checking the derivative on the entire interval. For logarithms, forgetting the factor \(2\), using a denominator that does not match the power, or treating a positive tail as if it could have either sign can also invalidate the conclusion.

These methods matter whenever a function value must be estimated with controlled precision. The useful result is not simply a string of digits: it is a numerical estimate paired with a reason those digits are reliable.

Check Your Understanding

For each question, identify what expansion or error information you would use, and give a justified conclusion where requested.

  1. For \(e^{0.1}\), what derivative bound would you use for a degree-four Maclaurin polynomial on the interval from \(0\) to \(0.1\)?
  2. Write the degree-five Maclaurin polynomial for \(\sin(0.3)\), and identify the first omitted term used in an alternating-series error bound.
  3. For \(\ln(1.1)\), calculate \(z=(x-1)/(x+1)\). Why is this substitution suitable for the logarithmic series?
  4. If an estimate is \(0.73642\) with error at most \(0.00003\), is \(0.736\) justified to three decimal places? Explain using an interval.
  5. State the two conditions needed to use the first omitted term as an alternating-series error bound.

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